A false reasoning for a better intuition behind the universal coefficient theorem












2












$begingroup$


Let $X$ be a topological space. I am trying to understand the natural homomorphism
$$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.



The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.



Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
$$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.



Let



$$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
and
$$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$




  1. Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.

  2. Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
    $$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
    which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.


Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.



Could someone tell me where this reasoning fails ? Thanks a lot !






homology-cohomology







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $X$ be a topological space. I am trying to understand the natural homomorphism
    $$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
    induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.



    The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.



    Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
    $$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
    Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.



    Let



    $$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
    and
    $$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$




    1. Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.

    2. Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
      $$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
      which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.


    Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.



    Could someone tell me where this reasoning fails ? Thanks a lot !






    homology-cohomology







    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Let $X$ be a topological space. I am trying to understand the natural homomorphism
      $$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
      induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.



      The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.



      Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
      $$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
      Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.



      Let



      $$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
      and
      $$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$




      1. Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.

      2. Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
        $$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
        which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.


      Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.



      Could someone tell me where this reasoning fails ? Thanks a lot !






      homology-cohomology







      share|cite|improve this question









      $endgroup$




      Let $X$ be a topological space. I am trying to understand the natural homomorphism
      $$H^*(M,mathbb{Z}) to H^*(M,mathbb{R}),$$
      induced by the inclusion $mathbb{Z} hookrightarrow mathbb{R}$.



      The universal coefficient theorem says that this homomorphism may not be injective, namely when torsion elements exist. I am trying to understand this in a very elementary way, and to that aim I have written a false argument of injectivity. I would like to understand where this argument fails, so I could have more intuition behind this non-injectivity. Any help would be appreciated.



      Denote $C_k(M)$ the free abelian group generated by $k$-singular chains on $M$, and by:
      $$C^k(M,mathbb{Z}) := Hom(C_k(M),mathbb{Z}), quad C^k(M,mathbb{R}) := Hom(C_k(M),mathbb{R}).$$
      Then the boundary operator $partial$ on $C_k(M)$ dualizes to boundary operators $delta_{mathbb{Z}}$ and $delta_{mathbb{R}}$ on the two above groups respectively.



      Let



      $$Z^k_{mathbb{Z}} := ker(delta_{mathbb{Z}} : C^k(M,mathbb{Z}) to C^{k+1}(M,mathbb{Z})), quad Z^k_{mathbb{R}} := ker(delta_{mathbb{R}} : C^k(M,mathbb{R}) to C^{k+1}(M,mathbb{R})),$$
      and
      $$B^k_{mathbb{Z}} := im(delta_{mathbb{Z}} : C^{k-1}(M,mathbb{Z}) to C^k(M,mathbb{Z})), quad B^k_{mathbb{R}} := im(delta_{mathbb{R}} : C^{k-1}(M,mathbb{R}) to C^k(M,mathbb{R})).$$




      1. Since $Hom(M,-)$ is left exact, the inclusion $mathbb{Z} hookrightarrow mathbb{R}$ yields an injection $C^k(M,mathbb{Z}) hookrightarrow C^k(M,mathbb{R})$.

      2. Since $delta_{mathbb{R} | C^k(M,mathbb{Z})} = delta_{mathbb{Z}}$, we have inclusions
        $$B^k_{mathbb{Z}} subset B^k_{mathbb{R}}, quad Z^k_{mathbb{Z}} subset Z^k_{mathbb{R}},$$
        which yield a natural homomorphism $rho : H^k(M,mathbb{Z}) to H^k(M,mathbb{R})$ in cohomology.


      Suppose that $alpha in Z^k_{mathbb{Z}}(M)$ is such that $rho([alpha]) = 0$. This means that $alpha in B^k_{mathbb{R}}$. But $alpha in C^k(M,mathbb{Z})$, so $alpha in B^k_{mathbb{R}} cap C^k(M,mathbb{Z}) = B^k_{mathbb{Z}}$, and therefore $[alpha] = 0$.



      Could someone tell me where this reasoning fails ? Thanks a lot !






      homology-cohomology




      homology-cohomology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 27 '18 at 16:51









      BrianTBrianT

      1916




      1916






















          1 Answer
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          The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.



          For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:03










          • $begingroup$
            @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:06










          • $begingroup$
            Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:15








          • 1




            $begingroup$
            Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:20






          • 1




            $begingroup$
            Thanks a lot for your help.
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:22











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          active

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          $begingroup$

          The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.



          For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:03










          • $begingroup$
            @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:06










          • $begingroup$
            Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:15








          • 1




            $begingroup$
            Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:20






          • 1




            $begingroup$
            Thanks a lot for your help.
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:22
















          1












          $begingroup$

          The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.



          For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:03










          • $begingroup$
            @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:06










          • $begingroup$
            Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:15








          • 1




            $begingroup$
            Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:20






          • 1




            $begingroup$
            Thanks a lot for your help.
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:22














          1












          1








          1





          $begingroup$

          The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.



          For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.






          share|cite|improve this answer











          $endgroup$



          The error is at the end: the intersection of real boundaries and integral chains is not equal to the integral boundaries.



          For instance if you consider a chain complex $mathbb Z overset{2}to mathbb Z$ and change coefficients to $mathbb R$ then in degree 0 degree 1 one has $B^1_{mathbb R} cap C^1_{mathbb Z} = mathbb Z$ but $B^1_{mathbb Z} = 2mathbb Z$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 27 '18 at 17:11

























          answered Dec 27 '18 at 17:00









          BenBen

          4,283617




          4,283617












          • $begingroup$
            Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:03










          • $begingroup$
            @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:06










          • $begingroup$
            Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:15








          • 1




            $begingroup$
            Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:20






          • 1




            $begingroup$
            Thanks a lot for your help.
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:22


















          • $begingroup$
            Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:03










          • $begingroup$
            @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:06










          • $begingroup$
            Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:15








          • 1




            $begingroup$
            Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
            $endgroup$
            – Ben
            Dec 27 '18 at 17:20






          • 1




            $begingroup$
            Thanks a lot for your help.
            $endgroup$
            – BrianT
            Dec 27 '18 at 17:22
















          $begingroup$
          Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:03




          $begingroup$
          Thanks. Could you extend your comment ? What would be the "difference" between these two sets ?
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:03












          $begingroup$
          @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
          $endgroup$
          – Ben
          Dec 27 '18 at 17:06




          $begingroup$
          @BrianT I added an example - maybe that helps? If it’s not clear please let me know.
          $endgroup$
          – Ben
          Dec 27 '18 at 17:06












          $begingroup$
          Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:15






          $begingroup$
          Thanks a lot. So if I understand well, you identify $mathbb{Z} =C^{-1}(M,mathbb{Z})$ and $mathbb{Z} = C^0(M,mathbb{Z})$ ? Do you have an intuition of why these sets are different ?
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:15






          1




          1




          $begingroup$
          Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
          $endgroup$
          – Ben
          Dec 27 '18 at 17:20




          $begingroup$
          Basically if you rescale a linear map over the integers, it shrinks the image. But if you then pass to the rationals/reals, now division is possible and all rescalings must have the same image. So if you intersect them with the integer values, they all get the same set - ignoring the shrinking image. So, you can’t determine the integer boundaries from the rational/real boundaries alone.
          $endgroup$
          – Ben
          Dec 27 '18 at 17:20




          1




          1




          $begingroup$
          Thanks a lot for your help.
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:22




          $begingroup$
          Thanks a lot for your help.
          $endgroup$
          – BrianT
          Dec 27 '18 at 17:22


















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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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